Let $a,b,c$ be non-negative reals, and $k$ is the best possible constant Prove the inequality Let $a,b,c$ be nonnegative  real numbers. Prove that
$$a^3+b^3+c^3-3abc\geq k|(a-b)(b-c)(c-a)|$$
where $k=\left(\frac{27}{4}\right)^{1/4}(1+\sqrt{3})$ and that $k$ is the best possible constant.
Not sure how to go about proving this inequality, I have not dealt with a type of inequality that includes a best possible constant $k$.
Any help to solve and understand this would be greatly appreciated.
 A: The problem amounts to minimizing:
$$ R(a,b,c)= \frac{a^3+b^3+c^3 - 3 a bc }{(b-a)(c-b)(c-a)} $$
on the set $0\leq a <b< c$.
The directional derivative:
$$ \frac{d}{dt}_{|t=0} R(a+t,b+t,c+t)= \frac{3}{2}\frac{(b-a)^2+(c-b)^2+(c-a)^2 }{(b-a)(c-b)(c-a)} $$ is non-negative so any minimum must occur for $a=0$. By homogeneity we may assume $c=1$. So we are reduced to minimizing for  $0<b<1$:
$$ R(0,b,1)=\frac{1+b^3}{(1-b)b} .$$
And this happens (set derivative=0)  when 
$$0=b^4-2b^3-2b+1=(b^2-(1+\sqrt{3})b+1)(b^2-(1-\sqrt{3})b+1)$$
Only the first factor has real roots from which:
$$ b=\frac{1+\sqrt{3}}{2}- \sqrt{\frac{\sqrt{3}}{2}} \ \ \left( \ \mbox{and}
\ \ \ \frac{1}{b}=\frac{1+\sqrt{3}}{2}+ \sqrt{\frac{\sqrt{3}}{2}} \right)$$
Inserting into the previous it yields (after reduction) the cited constant $k$. To see this note that:
 $$ (1-b) \left[ (1-b)(1+b^2) \right] =(1-b)^2(1+b^2) = 1-2b+2b^2-2b^3+b^4 = 2 b^2$$
(use that $b$ is the root of the 4'th degree polynomial above).
Using also $1+b^3=(1+b)(1-b+b^2)=(1+b) \sqrt{3} b$ we get for the minimal value:
$$  \frac{1+b^3}{b(1-b)} =  \frac{(1-b)(1+b^2) (1+b)\sqrt{3}}{2 b^2}=\frac{\sqrt{3}}{2}\left( \frac{1}{b}+b\right) \left( \frac{1}{b} - b\right)=\sqrt{3} \left( 1+ \sqrt{3}\right) \sqrt{\frac{\sqrt{3}}{2}}$$
A: Left and right expressions are non-negative (AM-GM).
Adding $X$ to each variable does not change the right side but the left side increases by $3(a^2 + b^2 + c^2 - ab - bc - ca)X$ which has the same sign as $X$ (unless $a=b=c$ in which case the inequality holds for any $k$).  
Therefore, taking $X$ negative to shift all three variables downward makes the inequality sharper and the critical cases are when $abc=0$.  Given that reduction, the rest is routine computation.
Let $c=0$.  We now want to know which $k$ have
$a^3 + b^3 \geq k|ab(a-b)|$.
This is symmetric and homogeneous so WLOG take $b=1$ and $a > b$.  The optimum $k$ is the minimum of $K(t) = (1+t^3)/t(t-1)$ for $t>1$, which is $\sqrt{9 + 6 \sqrt{3}}$, equal to the value given in the question.
There is a pair of equal minima because $K(t) = K(1/t)$, from the symmetry of $a$ and $b$.
A: Without loss of generality, let $a\geq b\geq c\geq 0$.  Write $x:=a-b$ and $y:=b-c$.  Then, we are to find the largest $k\geq 0$ such that $$(x+2y+3c)\left(x^2+xy+y^2\right)\geq k\,xy(x+y)\,.$$
Note that we need
$$(x+2y)\left(x^2+xy+y^2\right) \geq k\,xy(x+y)$$
for all $x,y\geq 0$ (by taking $c=0$).  If $x=0$ or $y=0$, then the inequality above is trivial for any $k\geq 0$.  By setting $t:=\frac{x}{y}$ when $x,y>0$, we see that $k$ must satisfy 
$$k\leq \frac{(t+2)\left(t^2+t+1\right)}{t(t+1)}$$
for all $t>0$.  That is, the largest $k$ would be
$$k_\max=\inf\Biggl\{\frac{(t+2)\left(t^2+t+1\right)}{t(t+1)}\,\Big|\,t>0\Biggr\}\,.$$ The rest should be easy.
